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If \(\frac{t}{u} = \frac{x}{y}\) and \(\frac{t}{y} = \frac{u}{x}\) and \(t\), \(u\), \(x\), and \(y\) are nonzero integers, which of the following is true? A. \(\frac{t}{u}=1\) B. \(\frac{y}{x}=1\) C. \(t = u\) D. \(t = \pm u\) E. None of the above
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16 Sep 2014, 00:27
Official Solution:If \(\frac{t}{u} = \frac{x}{y}\) and \(\frac{t}{y} = \frac{u}{x}\) and \(t\), \(u\), \(x\), and \(y\) are nonzero integers, which of the following is true? A. \(\frac{t}{u}=1\) B. \(\frac{y}{x}=1\) C. \(t = u\) D. \(t = \pm u\) E. None of the above Given that: \(\frac{t}{u} = \frac{x}{y}\) and \(\frac{t}{y} = \frac{u}{x}\). So, \(\frac{t}{u} = \frac{x}{y}\) and \(\frac{t}{u} = \frac{y}{x}\) (from 2), which means that \(\frac{t}{u}\) and \(\frac{x}{y}\) equal to their reciprocals: \(\frac{t}{u}=\frac{u}{t}\) and \(\frac{x}{y}=\frac{y}{x}\). So, \(t^2=u^2\) and \(t^2=u^2\), which gives \(t=u\) (or which is the same \(t = \pm u\)) and \(x=y\) (or which is the same \(x = \pm y\)). Answer: D
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Re: M0616
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13 Feb 2015, 22:18
Eqn 1....Let t/u = x/y = k
Therefore t = uk, x = yk.....
Substitute in eqn 2...t/y = u/x
uk/y = u/yk
k^2 = 1
k = +/1
so t = +/ u



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Re: M0616
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17 Feb 2015, 11:55
Bunuel In questions where they ask which is true.. are they asking for must be true or which can be true? Asking since i solved this question as which can be true and stopped after choosing A as A was coming as true for a case where all numbers were equal.



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17 Feb 2015, 11:57
qw1981 wrote: Bunuel In questions where they ask which is true.. are they asking for must be true or which can be true? Asking since i solved this question as which can be true and stopped after choosing A as A was coming as true for a case where all numbers were equal. Which is true = which must be true.
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Re: M0616
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10 Jun 2015, 11:54
Why are we crossmultiplying ? The question does not state that integers t,x,u,v are positive...



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10 Jun 2015, 11:56
Randude wrote: Why are we crossmultiplying ? The question does not state that integers t,x,u,v are positive... It does not matter for equations, it does only for inequalities.
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Re: M0616
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11 Dec 2015, 05:31
Hi,
I did the following:
From (1), ty=xu (cross multiplying) From (2), tx=uy
so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +u)
What am I doing wrong?



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26 Apr 2016, 18:39
m9338 wrote: Hi,
I did the following:
From (1), ty=xu (cross multiplying) From (2), tx=uy
so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +u)
What am I doing wrong? t=u would be true for negative and positive values of u. But I can see that some people would choose E.



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Re: M0616
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15 Jun 2016, 02:34
m9338 wrote: Hi,
I did the following:
From (1), ty=xu (cross multiplying) From (2), tx=uy
so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +u)
What am I doing wrong? One explanation that i can offer here is as follows  The equation t(x+y)=u(x+y) holds good in almost all case, however (given x,y are non zero integers), consider a scenario where x= y. In this case you cannot say that t=u as you cant divide both sides with (x+y) !!! why? coz its 0 Incidentally this is one of the wrong answer choices as well
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Re: M0616
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18 Feb 2017, 07:53
I still can't understand why option D is correct, even though the question is MUST BE TRUE. Here is my thinking:
Let's say t, u, x and y are all equal to 1 (nonzero integer). Then according to option D: 1 = +1, which is only true for positive sign, but not negative. I mean the option D would be a MUST BE TRUE answer only if it was written like this: t = u OR t = u. As far as I understand t = +u means 't = u AND t = u'



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Re: M0616
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10 Jul 2018, 19:42
why is t = u the same thing as t = ±u? I don't understand that. Can someone please explain?



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10 Jul 2018, 20:27
gdhume wrote: why is t = u the same thing as t = ±u? I don't understand that. Can someone please explain? t = u means that the distance from t to 0 is the same as the distance from u from 0. So, either t = s, or t = s.
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Re: M0616
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13 Aug 2018, 07:24
m9338 wrote: Hi,
I did the following:
From (1), ty=xu (cross multiplying) From (2), tx=uy
so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +u)
What am I doing wrong? I still cannot understand why this is wrong. Could someone help me?



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Re: M0616
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05 Sep 2018, 06:40
So, t^2=u^2 and t^2=u^2 gives t=u
Could someone please elaborate the result derived above?



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05 Sep 2018, 06:45
Megha1119 wrote: So, t^2=u^2 and t^2=u^2 gives t=u
Could someone please elaborate the result derived above? By taking the square root. Remember: \(\sqrt{x^2}=x\). The point here is that square root function can not give negative result: wich means that \(\sqrt{some \ expression}\geq{0}\). So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to? Let's consider following examples: If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\). So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\). What function does exactly the same thing? The absolute value function! That is why \(\sqrt{x^2}=x\)
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Re: M0616
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07 Nov 2018, 10:26
Hi Bunuel and chetan2u, I am wondering if my approach is correct? Could you verify please? \(\frac{T}{U}\)=\(\frac{X}{Y}\) > \(TYXU=0\) \(\frac{T}{Y}\)=\(\frac{U}{X}\) > \(TXUY=0\) Combining the 2 equations: \(TYXU=TXUY\) \(TY+UY=TX+XU\) \(Y*(T+U)=X*(T+U)\) Can I correctly infern that either \(Y=X\) or \((T+U)=0\), therefore \(T=U\)? Thank you in advance!!!!
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Re: M0616
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07 Nov 2018, 10:37
m9338 wrote: Hi,
I did the following:
From (1), ty=xu (cross multiplying) From (2), tx=uy
so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +u)
What am I doing wrong? Hi, You will be right if (x+y) not equals to zero. But the stem says all are non zero integers that doesnt mean all are positive. so it is still possible that x is positive and y is negative with equal magnitude (or vice versa). Hence we can't just cancel out the x+y . Hope it helps. Kudos=Thanks



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Bunuel, Hi..I have a doubt in this problem. t/u=x/y and t/u=y/x which basically means that t/u is equal to x/y as well as its reciprocal (y/x) which only means that x and y has to be of the same value making t/u =1 (Option A). 2ndly, y/x=1 which also makes x/y = 1 therby t/u=x/y and also its reciprocal (y/x) thus option B also seems plausible. I understood your explanation for Option D but Id like to understand the gap in my reasoning if I were to choose either A or B. Thankyou



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01 Dec 2018, 07:01
yashna36 wrote: Bunuel, Hi..I have a doubt in this problem. t/u=x/y and t/u=y/x which basically means that t/u is equal to x/y as well as its reciprocal (y/x) which only means that x and y has to be of the same value making t/u =1 (Option A). 2ndly, y/x=1 which also makes x/y = 1 therby t/u=x/y and also its reciprocal (y/x) thus option B also seems plausible. I understood your explanation for Option D but Id like to understand the gap in my reasoning if I were to choose either A or B. Thankyou x/y = y/x does not mean that x = y. Crossmultiply: x^2 = y^2 > x = y. Which means that x and y are same distance apart from 0: x = y or x = y.
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